Question: What is the extraneous solution to these equations? $\dfrac{x^2 - 22}{x + 7} = \dfrac{-16x - 85}{x + 7}$
Answer: Multiply both sides by $x + 7$ $ \dfrac{x^2 - 22}{x + 7} (x + 7) = \dfrac{-16x - 85}{x + 7} (x + 7)$ $ x^2 - 22 = -16x - 85$ Subtract $-16x - 85$ from both sides: $ x^2 - 22 - (-16x - 85) = -16x - 85 - (-16x - 85)$ $ x^2 - 22 + 16x + 85 = 0$ $ x^2 + 63 + 16x = 0$ Factor the expression: $ (x + 7)(x + 9) = 0$ Therefore $x = -7$ or $x = -9$ At $x = -7$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -7$, it is an extraneous solution.